Talk:Non-prime solvable fundamental groups (Ex)

Suggestion: The first manifold is $M = S^3$$<wikitex>; Suggestion: The first manifold is $M = S^3$. Other than that, non-prime manifolds possess prime decompositions. So there must be at least two manifolds connected sum to each other. The fundamental group of $M$ is the free product of the fundamental groups of the various fundamental groups. Since it's solvable, note that each abelian quotient group can only contain elements from any individual fundamental group, otherwise it wouldn't be abelian. Consider the first time in the series $G_n \triangleright G_{n-1}$ such that $G_n$ contains an element $x$ from some $\pi_1(M_i)$, where no element from $\pi_1(M_i)$ appears in $G_{n-1}$. $xG_{n-1}x^{-1}$ is a contradiction to normality, since they are in a free product. </wikitex>M = S^3$. Other than that, non-prime manifolds possess prime decompositions. So there must be at least two manifolds connected sum to each other. The fundamental group of $M$$M$ is the free product of the fundamental groups of the various fundamental groups. Since it's solvable, note that each abelian quotient group can only contain elements from any individual fundamental group, otherwise it wouldn't be abelian. Consider the first time in the series $G_n \triangleright G_{n-1}$$G_n \triangleright G_{n-1}$ such that $G_n$$G_n$ contains an element $x$$x$ from some $\pi_1(M_i)$$\pi_1(M_i)$, where no element from $\pi_1(M_i)$$\pi_1(M_i)$ appears in $G_{n-1}$$G_{n-1}$. $xG_{n-1}x^{-1}$$xG_{n-1}x^{-1}$ is a contradiction to normality, since they are in a free product.